Question

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Answer 1

Question:- A man wearing a bullet proof vest stands on roller skates . The total mass is 80 kg . A bullet of mass 20 g is fired at 400 m / s . It is stopped by the vest and falls to the ground . What is then the velocity of the man ?

Answer: 0.1 m/s

Step by Step Solution:

Since initially man was standing in rest thus his initial velocity will be zero. After the strike bullet stops thus final velocity of bullet will be zero

Thus we have,

Mass of Man(m₁) = 80 kg

Mass of Bullet (m₂) = 20g = 0.02kg

Initial Velocity of Man (u₁) = 0

Initial Velocity of Bullet (u₂) = 400m/s

Final Velocity of Bullet (v₂) = 0

Let the final velocity of man after the strike be v₁

We know that by law of conversation of momentum, total momentum before strike will be equal to total momentum after the strike.

Thus,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(80×0) + (0.02×400) = (80×v₁) + (0.02×0)

0 + 8 = 80v₁ + 0

v₁ = 8/80

v₁ = 0.1 m/s

Thus the velocity of man after the strike will be 0.1m/s

Answer 2

Explanation:

$\large{\underline{\bf{ <u>Question</u>:-}}}$

A man wearing a bullet proof vest stands on roller skates . The total mass is 80 kg . A bullet of mass 20 g is fired at 400 m / s . It is stopped by the vest and falls to the ground . What is then the velocity of the man ?

Explanation:−

$\large{\underline{\bf{\green{Given:-}}}}$

A bullet of mass $M_1=20g=\frac{20}{1000}g$

Initial velocity of bullet $u=400m/s$

Mass of vest $M_2=80kg$

Final velocity of man(v)=?

$\huge{\underline{\bf{\red{Solution:-}}}}$

$\texttt{Initial momentum of bullet=M_1 \times u}$

$=\frac{20}{1000} \times 400 \\ \\ = 0.002 \times 400 \\ \\ = 8 \: kgm/s$

$\texttt{Final momentum of Man=M_2\times v}$

$=80 \times v$

$\large{\underline{\bf{\green{Applying\: law \:of \:conservation\: of\: momentum}}}}$

$\texttt{M_1 \times u=M_2 \times v}$

$8 = 80 \times v \\ \\ v = \frac{8}{80} \\ \\ v = 0.1m/s$

$\texttt{\fbox{This velocity of man after striking will be 0.1 m/s}}$